# 给定一个链表，旋转链表，将链表每个节点向右移动 k 个位置，其中 k 是非负数。 
# 
#  示例 1: 
# 
#  输入: 1->2->3->4->5->NULL, k = 2
# 输出: 4->5->1->2->3->NULL
# 解释:
# 向右旋转 1 步: 5->1->2->3->4->NULL
# 向右旋转 2 步: 4->5->1->2->3->NULL
#  
# 
#  示例 2: 
# 
#  输入: 0->1->2->NULL, k = 4
# 输出: 2->0->1->NULL
# 解释:
# 向右旋转 1 步: 2->0->1->NULL
# 向右旋转 2 步: 1->2->0->NULL
# 向右旋转 3 步: 0->1->2->NULL
# 向右旋转 4 步: 2->0->1->NULL 
#  Related Topics 链表 双指针 
#  👍 441 👎 0


# leetcode submit region begin(Prohibit modification and deletion)
# Definition for singly-linked list.
class ListNode:
    def __init__(self, val=0, next=None):
        self.val = val
        self.next = next
class Solution:
    def rotateRight(self, head: ListNode, k: int) -> ListNode:
        # if not head:
        #     return None
        # if not head.next:
        #     return head
        #
        # # close the linked list into the ring
        # old_tail = head
        # n = 1
        # while old_tail.next:
        #     old_tail = old_tail.next
        #     n += 1
        # old_tail.next = head
        #
        # new_tail = head
        # for i in range(n - k % n - 1):
        #     new_tail = new_tail.next
        # new_head = new_tail.next
        #
        # new_tail.next = None
        #
        # return new_head
        '''快慢指针'''
        if not head or not head.next or not k: 
            return head
        tmp, count = head, 0
        while tmp:
            tmp = tmp.next
            count += 1
        k = k % count
        if k == 0:
            return head
        fast = slow = head
        for i in range(k):
            fast = fast.next
        while fast.next:
            fast = fast.next
            slow = slow.next
        newHead = slow.next
        slow.next = None
        fast.next = head
        return newHead
# leetcode submit region end(Prohibit modification and deletion)
